In a study of 250 adults, the average heart rate was 70 beats per minute. Assume the population of heart rates is known to be approximately normal, with a standard deviation of 12 beats per minute. What does a margin of error of 2 for the 98% confidence interval of the average beats per minute mean?
There is a 1% chance that the population mean is less than 70 beats per minute. There is a 1% chance that the population mean is more than 70 beats per minute. There is a 99% chance that the population mean is between 58 and 82 beats per minute.
There is a 99% chance that the population mean is between 68 and 72 beats per minute.

Form the confidence level (99%), we calculate the "alpha" value

alpha=1-(99/100)=0.01

Now, calculate the critical probability (p-start)

p*=1-(alpha/2)=1-(0.01/2)=0.995

z-score could be calculated from p*,,,

z-score=2.57583

Now, calculate the Margin of Error:

standard deviation = s

number of adults = n

ME=(z*s)/(sqrt n)=(2.57583*12)/(sqrt 250) = 1.955

The confidence interval = mean +or- ME

                                     = 70+1.955  or  70-1.955

So that we can conclude that the 99% confidence interval for the mean beats per minute is 68.045 and 71.955

I hope you got the idea!

step-by-step explanation:

what is the figure? attach a picture.